Riemann tensor in 3d Cartesian coordinates

[Orodruin]

but the correct type of coordinate system

You inherently seem to think of a manifold in terms of its embedding in a higher dimensional space. You need to get rid of this prejudice and start considering them from their intrinsic properties only. From the manifold point of view, there are not different types of coordinate systems - a coordinate system is just a continuous bijective map from a part of the manifold to a subset of R^n. You can describe the manifold using whatever coordinates you like that satisfy this. But again - get rid of the embedding thinking.

 

[PeterDonis]

it appears we don't need just to use the correct number of coordinates, but the correct type of coordinate system.

This is not correct. You can use any coordinates you want. You just need to use the correct expression for the metric for the coordinates you are using.

Let's consider a very simple example: the 2-d Euclidean plane. You can use Cartesian coordinates $x, y$ on this plane; if you do, the metric is $ds^2 = dx^2 + dy^2$. Or you can use polar coordinates $r, \theta$; if you do that, the metric is $ds^2 = dr^2 + r^2 d\theta^2$. Both of these $ds^2$ expressions represent e, the same line element, just in different coordinates.

To illustrate with a concrete example, suppose we are considering the line element $ds^2$ between the origin and the point $0.1, 0.1$ in Cartesian coordinates. The squared length of this line element is $ds^2 = dx^2 + dy^2 = (0.1)^2 + (0.1)^2 = 0.02$. But if we are using polar coordinates, the same point is described by coordinates $0.1414, \pi/4$, and the squared length is $ds^2 = dr^2 + r^2 d\theta^2 = (0.1414)^2 + (0.1414)^2 * (0)^2 = 0.02$--the same squared length, because it's the same line element, just represented in different coordinates.

It would of course be quite wrong to take the Cartesian coordinate values and plug them into the metric in polar coordinates, or vice versa. You have to use the correct metric for the coordinates you pick; but as long as you do that, you can pick any coordinates you like.

 

[PeterDonis]

That's why the cylinder cannot be described by Cartesian coordinates.

There seems to be a terminology issue here. By "Cartesian coordinates" I think you mean coordinates in which the metric is $ds^2 = dx^2 + dy^2$ and the ranges of the coordinates are $- \infty < x < \infty$ and $- \infty < y < \infty$, with no periodicity in either coordinate (i.e., distinct values of each coordinate map to distinct points). But one can drop the second requirement without dropping the first; i.e., one can describe a cylinder using coordinates in which the metric is $ds^2 = dx^2 + dy^2$ but the $y$ coordinate, for example, is periodic, so $y = - L/2$ and $y = L/2$ describe the same point (if we fix the value of $x$).

 

[stevendaryl]

There seems to be a terminology issue here. By "Cartesian coordinates" I think you mean coordinates in which the metric is $ds^2 = dx^2 + dy^2$ and the ranges of the coordinates are $- \infty < x < \infty$ and $- \infty < y < \infty$, with no periodicity in either coordinate (i.e., distinct values of each coordinate map to distinct points). But one can drop the second requirement without dropping the first; i.e., one can describe a cylinder using coordinates in which the metric is $ds^2 = dx^2 + dy^2$ but the $y$ coordinate, for example, is periodic, so $y = - L/2$ and $y = L/2$ describe the same point (if we fix the value of $x$).

Yes, but some people require that a coordinate system must be a one-to-one map from an open set of $R^N$ to an open set of the manifold. If $(x,y) = (x,y+L)$, then it's not one-to-one. It's a picky point, but in the book that I read on differential geometry, there was that requirement.

 

[PeterDonis]

Since their coordinates have a difference of L, I don't see why these are the same point.

Because the $x$ coordinate in this chart is periodic; it acts like an angular coordinate (for example, the $\theta$ coordinate in cylindrical coordinates) in this respect.

one faces this same issue if one uses cylindrical coordinates

Yes, and that's fine. Coordinate charts in which a coordinate is periodic are perfectly fine.

we could correctly map points on the surface of a sphere using spherical coordinates.

Yes, you can. And you can also do so in "rectangular" coordinates (I'll use that term instead of "Cartesian" to avoid any confusion); but the metric will not be the same as the metric of a flat plane even if you call the coordinates $x$ and $y$. (For example, pervect mentioned Mercator coordinates on the 2-sphere; they are "rectangular" but the metric is not the same as that of a flat plane.)

Also, a 2-sphere has a property that a flat plane or a cylinder do not: a 2-sphere cannot be entirely covered by a single coordinate chart. Standard spherical coordinates do not cover the "poles", since there are coordinate singularities there. You can cover a 2-sphere with two charts (the simplest choices are stereographic projections around two antipodal points), but not with a single chart.

 

[PeterDonis]

If $(x,y)=(x,y+L)$, then it's not one-to-one.

Wouldn't that invalidate any chart with angular coordinates? For example, polar coordinates on a plane? I have never seen a chart like that labeled as invalid (but of course I have not delved very deeply into differential geometry texts written from a mathematician's rather than a physicist's point of view).

 

[stevendaryl]

Wouldn't that invalidate any chart with angular coordinates? For example, polar coordinates on a plane? I have never seen a chart like that labeled as invalid (but of course I have not delved very deeply into differential geometry texts written from a mathematician's rather than a physicist's point of view).

It's a picky point that nobody cares about, but I would say that polar coordinates aren't really valid for the entire plane. The most egregious problem is that the metric components are singular at the origin. The fact that $(r, \theta)$ and $(r, \theta + 2\pi)$ are identified is not very serious. Basically, using $\theta$ modulo $2\pi$ IS using multiple charts: You have one chart where $\theta$ goes from $0$ to $2\pi$, and another where $\theta$ goes from $-\pi$ to $\pi$, and the mapping is just the modulo mapping.

 

[PeterDonis]

I would say that polar coordinates aren't really valid for the entire plane. The most egregious problem is that the metric components are singular at the origin

I agree there is a coordinate singularity at the origin, but I'm not asking about that.

Basically, using $\theta$ modulo $2\pi$ IS using multiple charts

Hm, ok.

 

[davidge]

You inherently seem to think of a manifold in terms of its embedding in a higher dimensional space

a coordinate system is just a continuous bijective map from a part of the manifold to a subset of R^n. You can describe the manifold using whatever coordinates you like that satisfy this. But again - get rid of the embedding thinking.

Oh ok.

Because the xxx coordinate in this chart is periodic; it acts like an angular coordinate (for example, the θθ\theta coordinate in cylindrical coordinates) in this respect.

Yes, and that's fine. Coordinate charts in which a coordinate is periodic are perfectly fine.

Yes, you can. And you can also do so in "rectangular" coordinates (I'll use that term instead of "Cartesian" to avoid any confusion); but the metric will not be the same as the metric of a flat plane even if you call the coordinates xxx and yyy. (For example, pervect mentioned Mercator coordinates on the 2-sphere; they are "rectangular" but the metric is not the same as that of a flat plane.)

Thanks for the answers to these questions

You can use any coordinates you want. You just need to use the correct expression for the metric for the coordinates you are using.

I see

Let's consider a very simple example: the 2-d Euclidean plane

To illustrate with a concrete example, suppose we are considering the line element ds2ds2ds^2 between the origin and the point 0.1,0.10.1,0.10.1, 0.1 in Cartesian coordinates

Thanks Peter for introducing this example. If the two points were $(1 , 1)$ and $(2 , 1)$ instead of $(0 , 0)$ and $(0.1 , 0.1)$, what value would we put for $r$ in the $r^{2}(\Delta \theta)^2$ term? Or more generally, what value do we put for $r$ when the two points are very close together, but the $r^{2}(d\theta)^2$ term don't vanish?

 

[PeterDonis]

If the two points were $(1 , 1)$ and $(2 , 1)$ instead of $(0 , 0)$ and $(0.1 , 0.1)$, what value would we put for $r$ in the $r^{2}(\Delta \theta)^2$ term?

You would have to solve an integral, since (assuming the coordinates you give are Cartesian), both $r$ and $\theta$ are changing. The general form of the integral would be

$$
s = \int ds = \int \sqrt{ds^2} = \int \sqrt{dr^2 + r^2 d\theta^2}
$$

which is not solvable as it stands. To solve it, you would need either $\theta$ as a function of $r$, or $r$ as a function of $\theta$ (i.e., you need an equation of the curve along which you are integrating), so that you can reduce the integral to an integral over one variable.

 

[davidge]

Ah ok

Is it possible to calculate $s$ on the surface of a sphere using $\int_{A}^{B}dx^2 + dy^2$ with $dx$ and $dy$ as in the image below (where I've set $z = 0$ along the path)?

 

[Nugatory]

Ah ok

Is it possible to calculate $s$ on the surface of a sphere using $\int_{A}^{B}dx^2 + dy^2$ with $dx$ and $dy$ as in the image below (where I've set $z = 0$ along the path)?

Yes, although what you're calculating is the length of the curve in three-dimensional space using the metric for that three-dimensional space written in Cartesian coordinates ($dx^2+dy^2$ is just another way of writing $g_{xx}dx^2+g_{yy}dy^2+g_{zz}dz^2=g_{ij}x^ix^j$ when $dz$ is zero). That's a different calculation than the one you'd make for the surface of a two dimensional sphere that is not embedded in a three-dimensional euclidean space.

BTW, we're trying to get you to abandon the embedding because the four-dimensional spacetime of general relativity cannot be usefully embedded in a five-dimensional Euclidean space. Thus, you're going to have to learn the techniques of differential geometry without relying on embedding before you can move forward with GR.

 

[davidge]

Yes, although what you're calculating is the length of the curve in three-dimensional space using the metric for that three-dimensional space written in Cartesian coordinates

BTW, we're trying to get you to abandon the embedding because the four-dimensional spacetime of general relativity cannot be usefully embedded in a five-dimensional Euclidean space. Thus, you're going to have to learn the techniques of differential geometry without relying on embedding before you can move forward with GR.

Oh yea, now I understand that. I asked about the integral $\int_{A}^{B} dx^2 + dy^2$ just for curiousity. Yet, how could we evaluate it? I know we cannot just replace the $d's$ with $\Delta 's$ in the integral, because in this case we would have a straight line through the interior of the sphere, not the path through the surface of it.

Edit: oops, I know that we cannot just replace the $d's$ with $\Delta's$ in any integral (it is meaningless). What I meant was that the result would not be as just $\Delta x^2 + \Delta y^2$.

 

[Nugatory]

Yet, how could we evaluate it?

By far the easiest way to evaluate that integral is to transform into polar coordinates. Choose the origin and axes so that the curve lies in the equatorial plane so that $d\phi$ is zero, $dr$ is already zero, and then $ds^2=g_{ij}x^ix^j$ will simplify down to $ds=\sqrt{g_{\theta\theta}d\theta^2}=Rd\theta$ and the integral is trivial. (The integral also looks formally similar to the integral you'd do without relying on the embedding but using latitude and longitude coordinates on the surface of the sphere. That's a consequence of choosing those particular coordinate systems, not any fundamental truth about the geometry).

 

[davidge]

By far the easiest way to evaluate that integral is to transform into polar coordinates

Yea. The thing is that I would like to evaluate that integral without any dependence with another coordinate system. If I were to use polar coordinates on the integral, I would not even write down the metric in Cartesian coordinates.

 

[Nugatory]

Yea. The thing is that I would like to evaluate that integral without any dependence with another coordinate system. If I were to use polar coordinates on the integral, I would not even write down the metric in Cartesian coordinates.

If you want to take on that problem, you can...You'll find that integral in a table of integrals, and if you don't you could always use numerical methods to evaluate it. But why? Coordinate systems have no physical significance, so the only reason to use one that makes a problem unnecessarily difficult is because you enjoy the challenge. Physical insight doesn't come from brute-forcing your way through an awkward coordinate system, it comes from finding the coordinate system that makes a problem easy (Einstein's discovery of general relativity is the most spectacular example).

 

[Nugatory]

...because in this case we would have a straight line through the interior of the sphere, not the path through the surface of it.

But there's nothing wrong with that. You're working in a three-dimensional space to evaluate the length of a curve in that three-dimensional space so you're allowed to use points off the curve. The integration is a matter of drawing lines in that space that are not exactly on the curve but approximate it. That is an altogether different problem than evaluating the length of a curve on the surface of a two-dimensional sphere (although with proper choice of embedding you can make the two problems look very similar, and the answer comes out the same).

Throughout this thread I've been a bit unclear about what you're really looking for... I think you're looking for a way to work with the surface of a two-dimensional sphere when the points on the surface of the sphere are labelled with the same $x,y$ values that they have in a three-dimensional space in which the sphere has been embedded? If so, the following definition of coordinates on the surface of the sphere will do the trick:
$x=X$
$y=Y$
$z=\sqrt{R^2-X^2-Y^2}$
(I've used $X,Y$ for the coordinates on the surface of the sphere to avoid any ambiguity about when I'm referring to the mapping from points on the surface of the sphere onto $\mathbb{R}^2$ and when I'm referring to the mapping between points in the three-dimensional space onto $\mathbb{R}^3$).

Plug these into the formula for the induced metric, just as you would the latitude/longitude coordinates I used in the earlier example, and you will get (but note my earlier caution about wise people checking my algebra!) the metric:
$g_{XX}=1+\frac{X^2}{R^2-X^2-Y^2}$
$g_{YY}=1+\frac{Y^2}{R^2-X^2-Y^2}$
$g_{XY}=g_{YX}=0$
(Strictly speaking, these coordinates only apply to a hemisphere, not the full sphere. That's not surprising because we already know that a two-dimensional sphere cannot be covered by a single coordinate patch. The coordinate singularity along the curve $X^2+Y^2=R^2$ is an indication of the problem).

 

[davidge]

If you want to take on that problem, you can...You'll find that integral in a table of integrals, and if you don't you could always use numerical methods to evaluate it. But why? Coordinate systems have no physical significance, so the only reason to use one that makes a problem unnecessarily difficult is because you enjoy the challenge. Physical insight doesn't come from brute-forcing your way through an awkward coordinate system, it comes from finding the coordinate system that makes a problem easy (Einstein's discovery of general relativity is the most spectacular example).

It's just to know if it's possible to do that.

Throughout this thread I've been a bit unclear about what you're really looking for... I think you're looking for a way to work with the surface of a two-dimensional sphere when the points on the surface of the sphere are labelled with the same $x$,$y$ values that they have in a three-dimensional space in which the sphere has been embedded?

Yes. This motivated me to start the thread, although the title suggests I'm most interested on the Riemman tensor.

the following definition of coordinates on the surface of the sphere will do the trick:
$x=X\\ y=Y\\ z=\sqrt{R^2−X^2−Y^2}$

In this case we could not evaluate it through the path for which $z = 0$?

Just to know... Would the integral look like this?

$$ \int ds = \int g_{XX}dX^2 + \int g_{YY}dY^2 = \int \int dX dX + \int \int {\frac{X^2}{R^2 - X^2 - Y^2}}dX dX + \int \int dY dY + \int \int {\frac{Y^2}{R^2 - X^2 - Y^2}}dY dY $$

 

[Nugatory]

In this case we could not evaluate it through the path for which z = 0?

There is no $z$ in the two-dimensional surface of the sphere, so there is no such thing as a path for which z=0 and the question is meaningless. If you're working in the three-dimensional euclidean space using x,y, and z coordinates then there are curves along which z=0 and you can use the integral from your post #46 to find the length along such a curve. But note the way you wrote the bounds of the integration in that post - it's a single line integral, not a double integral across dx and dy so...

Just to know... Would the integral look like this?

No. You would have to do something analogous to what PeterDonis suggested in #45, writing the values of $X$ and $Y$ along the curve as functions of a single integration variable so you can evaluate the line integral along the curve.

 

[Nugatory]

although the title suggests I'm most interested on the Riemman tensor.

And now that you have the metric in these (remarkably difficult to work with) coordinates, you can calculate the components of the Riemann tensor in these coordinates.

 

[davidge]

There is no zzz in the two-dimensional surface of the sphere, so there is no such thing as a path for which z=0 and the question is meaningless. If you're working in the three-dimensional euclidean space using x,y, and z coordinates then there are curves along which z=0 and you can use the integral from your post #46 to find the length along such a curve

I used the points $(0\ , 5\ , 0)$ and $(1\ , \sqrt{24}\ , 0)$ and the relation $x^2 + y^2 = 25 = R^2$, $d\phi = dr = 0$, such that the line element in sph. coord. is $ds^2 = R^2 d\theta^2$.
$$ S = \int_{5}^{\sqrt{24}} dy \sqrt{1+ \frac{y^2}{R^2 - y^2}} = \int_{arcos(0.4)}^{\frac{\pi}{2}} R d \theta $$
The result I got was the same for the two integrals above.

And now that you have the metric in these (remarkably difficult to work with) coordinates, you can calculate the components of the Riemann tensor in these coordinates

I will work on it

 

[davidge]

I found two non-vanishing components of the Riemann tensor, namely
$$R^{y}\ _{xxy} = \frac{(2y^4/x^2)+3y^2}{(x^2+y^2)^2} = -R^{y}\ _{xyx}$$
Is it correct?

 

[PeterDonis]

Would the integral look like this?

No. What you wrote is an integral for $ds^2$, not $ds$. But that integral doesn't make sense, because you are integrating along a line element so you need to have just one integration variable, whereas you have two. In other words, you've written what's supposed to be the integral along a curve, but you're treating it as though it were an integral to determine an area.

The correct general form for the integral of the line element along a curve would be

$$
\int ds = \int \sqrt{ds^2} = \int \sqrt{g_{XX} dX^2 + g_{YY} dY^2}
$$

Which can't even be evaluated as it stands; you would need to find an equation for the curve, i.e., for $X$ in terms of $Y$ or $Y$ in terms of $X$ (or both of them in terms of some curve parameter), and use that equation to get the integrand into a form where it is a valid integrand with one integration variable.

 

[davidge]

@PeterDonis I constructed an integral following these steps and it worked. Please see my post #56.

 

[PeterDonis]

I constructed an integral following these steps and it worked. Please see my post #56.

Ah, ok, got it.

 

[PeterDonis]

using $\theta$ modulo $2\pi$ IS using multiple charts

I was hesitant in my response before but didn't have time to pin it down. Now I have; a while back we had a discussion about covering $S^1 \times R$ with a single chart, in which you posted:

The entire $S^1$ is an open interval, since it is the union of the interval $0 < \theta < \pi$ and $\frac{\pi}{2} < \theta < \frac{5 \pi}{2}$.

So it's a weird fact that $S^1$ cannot be covered by a single chart, but #S^1 \times R## can be.

This seems to suggest that we don't need to use multiple charts to have a coordinate, $\theta$ in this case, that is periodic.

 

[Nugatory]

I found two non-vanishing components of the Riemann tensor, namely
$$R^{y}\ _{xxy} = \frac{(2y^4/x^2)+3y^2}{(x^2+y^2)^2} = -R^{y}\ _{xyx}$$
Is it correct?

It's not promising that these expressions blow up at $X=Y=0$ where the coordinates and the manifold are well-behaved. Did you get these out of mathematica or equivalent, or do it by hand?

 

[Orodruin]

I was hesitant in my response before but didn't have time to pin it down. Now I have; a while back we had a discussion about covering $S^1 \times R$ with a single chart, in which you posted:
This seems to suggest that we don't need to use multiple charts to have a coordinate, $\theta$ in this case, that is periodic.

Strictly speaking, a chart must be a 1-to-1 mapping from the manifold to an open subset of $\mathbb R^n$ where $n$ is the dimension of the manifold. The typical one-chart atlas of the cylinder is to map it to a plane with one point removed - but you cannot use polar coordinates on that plane - you need to use the usual coordinates, or the map will not be 1-to-1.

 

[davidge]

It's not promising that these expressions blow up at $X = Y = 0$ where the coordinates and the manifold are well-behaved

Remember that I'm considering points along the surface of the sphere for which $z = 0$. So $x^2 + y^2 = R^2$, where $R$ is the radius of the sphere, and $X = Y = 0$ is not a possibility. The above expression for the two mentioned components would read, say in terms of y, $$\frac{2y^4 / (R^2 - y^2) + 3y^2}{R^4}$$

Did you get these out of mathematica or equivalent, or do it by hand?

I just calculated by hand the Cristoffel symbols and its derivatives and used it in the equation for the componets.

 

[Orodruin]

Remember that I'm considering points along the surface of the sphere for which $z = 0$. So $x^2 + y^2 = R^2$, where $R$ is the radius of the sphere, and $X = Y = 0$ is not a possibility. The above expression for the two mentioned components would read, say in terms of y, $$\frac{2y^4 / (R^2 - y^2) + 3y^2}{R^4}$$

I just calculated by hand the Cristoffel symbols and its derivatives and used it in the equation for the componets.

You are still doing it wrong then. You have to dismiss the idea of using the coordinates of the embedding space. On the part of the sphere where $z = 0$ in the embedding the coordinates $x$ and $y$ do not form a good coordinate system! You cannot use them as coordinates and compute stuff such as the components of the Riemann tensor, e.g., $R_{xyx}^y$.

 

[davidge]

the coordinates $x$ and $y$ do not form a good coordinate system

What do you mean by good coordinate system?

 

[Orodruin]

What do you mean by good coordinate system?

A continuous bijection from an open subset of $\mathbb R^2$ to a subset of the sphere.

 

[davidge]

A continuous bijection from an open subset of $\mathbb {R}^2$ to a subset of the sphere.

I don't see why it is not a bijection. Bijection means that each element on $\mathbb {R}^2$ is mapped on each point on the sphere, and no two points will be assigned to the same point on $\mathbb {R}^2$. Also, all points on the sphere will be mapped. Right? But the coordinates $x,y$ also do this job.

 

[stevendaryl]

I was hesitant in my response before but didn't have time to pin it down. Now I have; a while back we had a discussion about covering $S^1 \times R$ with a single chart, in which you posted:

This seems to suggest that we don't need to use multiple charts to have a coordinate, $\theta$ in this case, that is periodic.

Wow. I don't remember what I was thinking at that time. But you can come up with a mapping from the cylinder to the plane minus one point:

Map $z,\theta$ to the point $(x,y)$ with $x = e^z cos(\theta)$, $y = e^z sin(\theta)$. Every point on the cylinder is given a pair of coordinates, and distinct coordinates correspond to distinct points on they cylinder. But there is no point on the cylinder corresponding to the point $x=y=0$.

What I don't remember is whether a chart has to be simply-connected or not, or whether it's enough that it be an open set.

 

[Orodruin]

I don't see why it is not a bijection. Bijection means that each element on $\mathbb {R}^2$ is mapped on each point on the sphere, and no two points will be assigned to the same point on $\mathbb {R}^2$. Also, all points on the sphere will be mapped. Right? But the coordinates $x,y$ also do this job.

No they do not. At least not on the intersection of the embedding of the sphere in $\mathbb R^3$ and the plane $z = 0$ -- which is exactly where you are trying to make them work. If you pick a coordinate $x$ then the value of $y$ is completely determined. If it was a bijection you would be able to pick any values of $x$ and $y$ and they should correspond to a unique point on the sphere.

I will repeat one more time: Please stop thinking in terms of the embedding into $\mathbb R^3$.